# Is 1 / ( x + 3 )^2 strictly greater than  2/(x-3)?

Oct 26, 2015

No (not always, at least).

#### Explanation:

If it was, then the answer to "For which values of $x$ the following disequality holds?"

$\frac{1}{x + 3} ^ 2 - \frac{2}{x - 3} > 0$

Should be "Always". Let's check it. Make the fractions have a common denominator, and you'll end up with

$\frac{x - 3}{{\left(x + 3\right)}^{2} \left(x - 3\right)} - \frac{2 {\left(x + 3\right)}^{2}}{{\left(x + 3\right)}^{2} \left(x - 3\right)} > 0$.

Now sum the numerator:

$\left(x - 3\right) - 2 {\left(x + 3\right)}^{2} = x - 3 - 2 \left({x}^{2} + 6 x + 9\right) =$

$= x - 3 - 2 {x}^{2} - 12 x - 18 =$

$= - 2 {x}^{2} - 11 x - 21 = - \left(2 {x}^{2} + 11 x + 21\right)$

Since $2 {x}^{2} + 11 x + 21$ has a negative discriminant, it has no solution, and so it is always positive. So, the fraction becomes

$- \frac{2 {x}^{2} + 11 x + 21}{{\left(x + 3\right)}^{2} \left(x - 3\right)}$

and we know that $2 {x}^{2} + 11 x + 21$ and ${\left(x + 3\right)}^{2}$ are always positive (the latter being a square). So, the sign of the whole fraction depends on the sign of $x - 3$, which obviously is positive after $3$ and negative before. Since there is a minus before the whole fraction, we have that the expression is positive only before $3$, and so the answer is no:

$\frac{1}{x + 3} ^ 2$ is not always greater than $\frac{2}{x - 3}$, but only before $3$.